To find the area bounded by the curve $y = 4 - x^2$ and the x-axis between $x = 1$ and $x = 6$, we need to calculate the definite integral of $y = 4 - x^2$ with respect to $x$ over the interval $[1, 6]$.

The formula for the area under the curve is:

$A = \int_{1}^{6} (4 - x^2) \, dx$

Step 1: Integrate the function

First, we'll compute the integral of $4 - x^2$:

$\int (4 - x^2) \, dx = 4x - \frac{x^3}{3} + C$

Step 2: Apply the limits of integration

Now, we evaluate the integral from 1 to 6:

$A = \left[ 4x - \frac{x^3}{3} \right]_{1}^{6}$

Substituting the upper limit ($x = 6$):

$A_6 = 4(6) - \frac{6^3}{3} = 24 - \frac{216}{3} = 24 - 72 = -48$

Substituting the lower limit ($x = 1$):

$A_1 = 4(1) - \frac{1^3}{3} = 4 - \frac{1}{3} = \frac{12}{3} - \frac{1}{3} = \frac{11}{3}$

Step 3: Calculate the total area

Now, subtract the value at $x = 1$ from the value at $x = 6$:

$A = -48 - \frac{11}{3} = \frac{-144}{3} - \frac{11}{3} = \frac{-155}{3}$

Since the area cannot be negative, we take the absolute value:

$A = \frac{155}{3} \, \text{square units}$

Thus, the area bounded by the curve and the x-axis is $\frac{155}{3}$ square units.

Would you like further details on any step or have any questions?

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Here are 5 related questions:

1. How do you interpret the meaning of a negative result in definite integrals?
2. What is the geometric significance of an area calculation under a curve?
3. How would you calculate the area if the curve were above and below the x-axis?
4. What is the difference between definite and indefinite integrals in terms of area?
5. How do you calculate the area between two curves?

Tip: Always check whether the function crosses the x-axis within the given limits, as it may affect the area calculation.